= Solution
Differentiation gives $f'(m)=2m(b+2dm^2)$. For $b>0$ the origin is the <global minimum>; for $b<0$, $m=\sqrt{-b/(2d)}$ has $f''=-4b>0$ and $f=-b^2/(4d)<f(0)$. Therefore, close enough to the transition for the quartic expansion to apply,
$$
\boxed{m_{\rm eq}(T)=\begin{cases}\sqrt{(T_c-T)/(2dT_c)},&T<T_c,\\0,&T\geq T_c.\end{cases}}
$$
The <order parameter> approaches zero continuously with <mean-field critical exponent> $\beta=1/2$: \b[the transition is continuous]. In the aligned phase a particular direction is chosen. The vector-spin <rotation> <symmetry> is spontaneously broken from $SO(3)$ to $SO(2)$, or from $O(3)$ to $O(2)$ including reflections. The manifold of <vector order parameter> directions is a <sphere>, so this is not just a two-state sign choice.<Statistical Field Theory by David Tong>[§4.2, $O(3)$ Heisenberg model and the continuous set of ordered directions]{ref}
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2016/paper-1-landau-ferromagnet.png]
{title=Quartic ferromagnetic free energy and the continuous equilibrium magnetization curve}
The plot uses illustrative $d=1/2$ and <temperature> measured in units of $T_c$. A unit-spin average must satisfy $m\leq1$; if the quartic model is used literally far below the transition, its constrained minimum is $\min\{1,\sqrt{(T_c-T)/(2dT_c)}\}$. In practice the near-transition expansion is not a controlled prediction of deep low-temperature saturation. Also, a finite zero-field <statistical ensemble> that averages all directions has $\langle\mathbf m\rangle=0$ even below $T_c$; the curve describes the magnitude in a selected ordered state, obtained by appropriate <boundary conditions> or the <thermodynamic limit> before removing a small <magnetic field>.
Back to article page
